Question

a cask contains a mixture of 49 litres of wine and water in the proportion 5:2. how much water must be added to it so that the ratio of wine to water may be 7:4 ?

Answer 1

6 liters water must be added to the mixture.

Explanation

Total amount of the mixture = 49 liters.

Ratio of wine and water in the mixture is 5 : 2

So, the amount of wine $=49*(\frac{5}{5+2}) = 49*\frac{5}{7}=35$

and the amount of water  $=49*(\frac{2}{5+2}) = 49*\frac{2}{7}=14$

Suppose,  $x$ liters water is added to this mixture.

So, the amount of water in the new mixture will be:  $(x+14)$ liters

Given that, the ratio of wine and water in the new mixture is 7 : 4 . That means........

$\frac{35}{x+14}=\frac{7}{4}\\ \\ 7(x+14)=35*4\\ \\ x+14= \frac{35*4}{7} =20\\ \\ x=20-14=6$

So, 6 liters water must be added to the mixture.

Answer 2

Answer:

The required amount of water added is x=6 liter.      

Step-by-step explanation:

Given : A cask contains a mixture of 49 liters of wine and water in the proportion 5:2.

To find : How much water must be added to it so that the ratio of wine to water may be 7:4 ?

Solution :

Total mixture = 49 liters

The ratio of wine and water is 5 : 2

Sum = 5+2=7

Total wine is $W=49\times \frac{5}{7}=35$

Total water is $w=49\times \frac{2}{7}=14$

Let x water liter should be added,

According to question,

$35:14+x=7:4$

$\frac{35}{14+x}=\frac{7}{4}$

Cross multiply,

$35\times 4=7\times (14+x)$

$140=98+7x$

$7x=140-98$

$7x=42$

$x=\frac{42}{7}$

$x=6$

Therefore, The required amount of water added is x=6 liter.